Heather has a coin that has a $60\%$ chance of showing heads each time it is flipped. She is going to flip the coin $5$ times. Let $X$ represent the number of heads she gets. What is the probability that she gets exactly $3$ heads? You may round your answer to the nearest hundredth. $P(X=3)=$
Solution: Without a fancy calculator Getting $3$ heads in $5$ flips means Heather needs to get $3$ heads and $2$ tails. For each flip, we know $P({\text{heads}})={60\%}$ and $P({\text{tails}})={40\%}$. Let's start by finding the probability of getting $3$ heads followed by $2$ tails: $P({\text{HHH}}{\text{TT}})=({0.6})^3({0.4})^2=0.03456$ This isn't our final answer, because there are other ways to get $3$ heads in $5$ flips (for example, HHTTH). How many different ways are there? We can use the combination formula to find how many ways there are to get $3$ heads in $5$ flips: $\begin{aligned} _n\text{C}_k&=\dfrac{n!}{(n-k)!\cdot k!} \\\\ _5\text{C}_3&=\dfrac{5!}{(5-3)!\cdot3!} \\\\ &=\dfrac{5 \cdot 4 \cdot \cancel{3 \cdot 2 \cdot 1}}{(2 \cdot 1) \cdot \cancel{3 \cdot 2 \cdot 1}} \\\\ &=10 \end{aligned}$ There are $10$ ways she can get $3$ heads in $5$ flips. Do they all have the same probability? Each of the $10$ ways has the same probability that we already found: $\begin{aligned} P({\text{H}\text{H}\text{H}}{\text{TT}})&=({0.6})^3({0.4})^2=0.03456 \\\\ P({\text{HH}}{\text{T}}{\text{H}}{\text{T}})&=({0.6})^3({0.4})^2=0.03456 \\\\ \vdots \\\\ P({\text{TT}}{\text{H}\text{H}\text{H}})&=({0.6})^3({0.4})^2=0.03456 \end{aligned}$ So we can multiply this probability by $10$ since that is how many ways there are to get $3$ heads in $5$ flips: $\begin{aligned} P(X=3)&=10(0.6)^3(0.4)^2 \\\\ &=10(0.03456) \\\\ &=0.3456 \\\\ &\approx 0.35 \end{aligned}$ Answer $P(X=3)=0.3456\approx0.35$